Batteries

CSM Engineer, i think you answer is right ! it is very best answer to this question !

the voltage is more higher ,but the time is not .the total voltage is number*voltage(one pcs), but the work time is not longer than one .of course ,above all is only my opinion for reference

Regards

yejic

Yes, more voltage and same current. And ofcourse more power(10 times more than single battery).

No, Holmes:

single battery: P₁=V²/R_L

10 batteries in series: P₂=V_total²/R_L=(10.V)²/R_L=100.V²/R_L=100.P₁

(Hence 100 times more than a single battery)

[Of course, when 10 batteries are connected in parallel, you have the same power as with a single battery (same voltage applied on the same load)... (And, of course, you have the same load current too... the only difference is that, in this case, each battery supplies only the 1/10 of the load current, thus the operation of the circuit will last 10 times longer (until the batteries become empty)...)...]

(Of course, we speak about "ideal" batteries, e.g. without internal resistance)...

thank you that was what i was thinking and i was unsure if i was right

so if you look at the pack as 10 x 1000 amps and you charge it eack battery will need 100 amps of charge to keep it in a good run condition. if i read this right ?

I'm afraid that I didn't get your question... Anyway, the charge of the batteries is another story... What I said in the previous post is just this: case1: suppose that you give I_L=10A to a load from just one battery... case2: suppose that you give, again, I_L=10A to this load from a a pack of "10 batteries connected in parallel" this time... in the case2 each battery supply only the 1/10 of the I_L, hence 1A ... You can conclude that the operation of the circuit (e.g. to supply I_L to the load) will last 10 times longer in case2 than in case1...

Note: Of course, in the above example (previous post) we assume that all the batteries have the same Ah (amperhour) value in both the case1 and case2...

Good points...I think...such as, when you reshuffle those series batteries, you get back some depleted voltage potential. ???

You said: ".....reshuffle those series batteries....." I suppose that you mean to change the sequence (placement) of the series batteries... Then you said: ".....you get back some depleted voltage potential. ???....." Of course not... This doesn't change anything... The total voltage supply is the sum of the voltages of each battery anyway (hence it is the same regardless of the sequence of the batteries)...

Then it must be something else which explains why reversing cells in a dimming flashlight (torch) will brighten the bulb. Don't believe? Try it.

Unbelievable... ...Have you observe such a thing in many cases???... (or just once???)... Have anyone else observe such a thing too???... ...

GK,

Mind you, I'm not saying the effect is long lasting. So often it seems, that when I need to use a flashlight after its having laid unused for a time, it has dimmed considerably from what I expected remembering its last use. It has been my practice for a very long time, and on innumerable occasions, to reverse the cells to get a temporary boost in brightness...not always a lot but often enough.

Memory fails at this point...but I supposed that, when I first observed it decades ago, my thinking might have been on the order of:

• Whatever release of ions occurs at one end of a cell must be accompanied, after work done, by a capture of ion at anode at the other end;
• And that as a cell weakened--became more resistant (after a manner of speaking) to internal transport of ions--then there must occur a kind of stratification within the cell, and
• if such is the case, were it possible, then if the stratification could somehow be reversed within a cell...(reader fills in the rest)...
• but that with multiple, stacked cells and
• said stratification occurring across the cells, both individually and as a unit, then
• indeed it might be possible (after a manner of speaking) to reverse the stratification by reversing (rearranging) the cells
• such that the somewhat higher availability of electrons at the cathode of the hindmost cell
• could, by placing that cathode closest to the load,
• result in a slight increase in voltage across the load...
• until such effect of stratification happened again with respect to the new arrangement of cells...
• until such time as both (or all) cells were effectively neutralized to the point of being dead...

Or it could have been that the experiment was done out of mindless curiosity and the explanation came afterwards. In either event, the phenomenon always occurs when I do the trick...provided there is enough, pre-cell-switch bulb illumination such that any change in bulb brightness would be perceivable after the switch. It would be my guess that the phenomenon would happen even after then--up to the point of total cells deadness--but would be imperceptable...as in, who could distinguish, say, a 1/50 volt delta by the brightness delta of a heated filament? (Conversely, with "new" cells, how difficult might it be to perceive the difference of brightness produced as between (say) 3.4 (2 x 1.7) and 3.34 (2 x 1.67) volts?

Curiously, it was just this afternoon that I had occasion to turn on a 4xD-cell lantern devise only to discover that the light was almost completely dimmed. (I would not learn until later that the lantern had be left On resulting in near depletion of the cell battery.) Not knowing whether it was defect of lantern or of battery (the recently replaced cells had been expected to persist until next June), but hoping for the latter, I opened and shuffled the cells. Not only did the bulb come back, it came back with an intensity I would judge to be not less that 60% of "new battery" intensity. That level of bulb brightness (minus perhaps 50%) persisted for about 5 minutes, at which point the bulb (suddenly) diminished steadily over about a five second period (as if controlled by a dimmer switch) before disappearing completely. (Subsequent battery cell renewal restored the lantern to full functioning.) In this case the remarkable voltage gain (as compared with common flashlight) by cell rearrangement might be explained by electronic component(s) within the lighting circuit; I have never disassembled the lantern to find out what....

Hello Guest,

Such a phenomenon would violate basic laws of physics.

Imagination is a wonderful thing.

Kind Regards....

phenomenon not imagined.

Remark not appreciated.

Hello again Guest,

Unless the torch has some electronic control unit inside, you would find the "brighter appearance" of the lamp, which you say was obtained by merely reversing the polarity of the batteries, was purely subjective, I'm sure.

However, I have thought of a possible reason for your "brighter lamp" situation.

Often in a dry cell situation, the contacts of each cell get a thin oxide film, and a thin oxide film sometimes builds up on the spring contacts.

So, if you just rotated the cells, and scratched through that thin oxide film, the bulb may have glowed brighter, if an oxide film had been the cause of slightly higher resistance contact/s in the series circuit.

If this oxide film was the cause, then the "brighter lamp" was not caused by the polarity reversal of the cells, but instead by the lessening of contact resistance, caused by movement/scraping.

Cheers and Kind Regards from far away....

Keep in mind though that with a series connection the max current will be limited by the internal impedance of each battery so if even one in the stack is partially charged that one, the weakest link, will limit your available output current.